Why order and repetition cannot make two sets different

0:00Here are two sets. One, two, three, four. And three, one, four, two. Same or different? Most people say same, and quickly. Ask why, and the answers get vague. So let us ask the harder question first. What could possibly make two sets different? Not the order, apparently. What about writing something twice? What about completely different wording? A set is fixed by one thing only: which objects answer yes to its membership question.

0:34Everything else - the order of the list, a repeated entry, the sentence you used - is a fact about the writing, not about the set. So the whole topic is one idea: two sets are the same set when they have the same members. Which gives us the test, and it has two halves. Take everything in the first set. Is each one in the second? Then turn around. Take everything in the second set. Is each one in the first?

1:05If both halves come back clean, the sets are equal. Two directions, and you have to run both. That matters more than it sounds. Notice what the test does not ask. Not whether the descriptions match, not whether the lists look alike. It asks the membership question, once per member, and takes the answers. So run it on the pair we opened with. One, two, three, four. And three, one, four, two.

1:35Is one in the second set? Yes. Is two? Yes. Three? Yes. Four? Yes. First direction clean. Now back. Is three in the first set? Yes. One? Yes. Four? Yes. Two? Yes. Second direction clean. Equal. And notice that at no point did the test ask where anything was sitting. It could not have. The membership question is: is this object in the set. There is nowhere in that question for a position to go.

2:07Order is something a list has. A set does not have one, because a set is not a list - it is the answer to a question. Now the same argument, applied to something that looks like a different problem. One, two, three. Against one, two, two, three, three. The second one writes two twice and three twice. Five entries against three. Run the test. Is one in the second set? Yes. Two? Yes. Three? Yes.

2:39Back the other way. Is one in the first? Yes. Two? Yes. Two again? Still yes - nothing has changed. Three? Yes. Three again? Yes. Both directions clean, so they are the same set. And notice why. Asking whether two is in the set a second time cannot produce a different answer. The question has one answer, so writing the object twice buys you nothing. A repeat is not a second member. It is the same member, mentioned again.

3:15Here is the case that breaks people's intuition, in both directions at once. First set: the prime numbers below six. Second set: the prime factors of thirty. Those two sentences have almost nothing in common. One is about a range, one about a single number. Work them. The primes below six are two, three and five. Thirty is two times three times five, so its prime factors are two, three and five.

3:46Same three. So the sets are equal. Which kills two misconceptions with one example. Sets are not equal because they are described the same way, and they are not unequal because they are described differently. The description is a route to the members. Two different routes can land in the same place. Now the mistake that costs the most marks, and it is not careless. Take one and two. Against one, two, three.

4:18Run the first half of the test. Is one in the second set? Yes. Is two? Yes. First direction completely clean. Every single member of the first set is in the second. Stop there and you would say equal. And you would be wrong. Turn around. Is one in the first set? Yes. Two? Yes. Three? No. Three is in the second set and not in the first. So they are not equal, and everything you did in the first direction was correct.

4:51That is what makes this one dangerous. One direction tells you the first set is contained in the second. Containment is a real relationship and it is not equality. Both halves, every time. The second half is where the extra member hides. Which raises the question of what it takes to prove two sets are not equal. Less than you think. To prove they ARE equal you need every member checked, both ways. There is no shortcut, because a single unchecked member could be the one that fails.

5:30To prove they are NOT equal you need exactly one member. One object that is in one set and not in the other. Name it, say which side it is on, say why it fails on the other side. Done. You do not have to list either set, or find all the differences. One is a complete proof. And that asymmetry is not a quirk of sets. Proving something holds everywhere is expensive; proving it fails somewhere costs one example.

6:04Five sets, and the job is to find every equal pair. A is the set holding zero. B is the numbers bigger than fifteen and smaller than five. C is given by x minus five equals zero. D is given by x squared equals twenty-five. And E is the positive whole-number solution of x squared minus two x minus fifteen equals zero. Work each one down to its members before comparing anything. That is the whole method.

6:39B first, because it pays off immediately. Nothing is both above fifteen and below five, so B is empty. And an empty set is unequal to every set with a member in it. So B is out against all four others at a stroke - each settled by naming any member of the other side. C is straightforward: x is five. So C holds five. D: x squared equals twenty-five gives five and minus five. Two members.

7:12E: that quadratic factors as x minus five times x plus three, so the solutions are five and minus three. Only five is a positive whole number, so E holds five. Now compare. A holds zero, so A fails against everything. C and D differ by minus five. But C holds five and E holds five, and neither holds anything else. Exactly one equal pair out of the five sets, and it is C and E - two conditions that look nothing alike.

7:49Two more pairs, and the second is worth arguing about. First: the letters of the word ALLOY, against the letters of the word LOYAL. ALLOY has five letters and four distinct ones: A, L, O and Y. LOYAL also has five letters, also doubles the L, and gives the same four. Different order, a doubled letter in each, and the same set. Both of the rules we proved, in one example.

8:19Second: the whole numbers, positive and negative, whose square is four or less. Against the solutions of x squared minus three x plus two equals zero. The first set: minus two, minus one, zero, one, two - every square four or less. The second: that quadratic factors as x minus one times x minus two, so it holds one and two. Not equal. And here is the thing worth saying out loud.

8:50Zero is in the first and not the second, so zero settles it. But so does minus two. And so does minus one. Three different members would each finish the argument on their own, and none of them is the right one, because there is no right one. A disproof needs a witness. It does not need a particular witness. Pick whichever you spot first. Here is one that catches almost everybody, and the trap is a minus sign.

9:22Is the set holding two and three the same as the solutions of x squared plus five x plus six equals zero? It looks like it should be. Five and six, two and three, two plus three is five and two times three is six. Everything lines up. So factor it. x squared plus five x plus six is x plus two times x plus three. Which means the solutions are minus two and minus three.

9:53Not two and three. Minus two and minus three. So the sets are unequal, and one member proves it: two is in the first set and is not a solution. The numbers you spot inside the quadratic are not its solutions. They add and multiply correctly, and the solutions are their negatives. Solve it. Do not recognise it. Eight sets now, to be sorted into families of equal ones. Two, four, eight, twelve. One, two, three, four. Four, eight, twelve, fourteen. Three, one, four, two. Minus one and one. Zero and a. One and minus one. Zero and one.

10:38Take the easy ones first. One, two, three, four and three, one, four, two are the same four numbers scrambled, so they are equal. Minus one and one, against one and minus one. Same two, scrambled. Equal. Now the first and third. They share four, eight and twelve. But the first has two, which the third does not, and the third has fourteen, which the first does not. Either one of those settles it. They are unequal, and three shared members out of four changes nothing.

11:15That leaves zero-and-a against zero-and-one, and this one needs a decision made out loud. Nobody said what a is. If a is some object we have not been told about, then it is not one, so the sets are unequal. If a were one, they would be equal. The verdict depends on which reading you take, and nothing picks one. So take the honest reading: an undefined letter is unknown, not convenient. They are unequal.

11:46Two families, then, and four sets that belong to no family at all. One more, and it is the shortest argument here. The letters needed to spell CATARACT, against the letters needed to spell TRACT. CATARACT is eight letters long. TRACT is five. So the first set is bigger. Obviously. No. CATARACT uses C, A, T and R - and then repeats them. Four distinct letters. TRACT uses T, R, A and C. The same four.

12:21Equal. An eight-letter word and a five-letter word naming one four-member set. The length of the word is a fact about the word. The set cannot know it, because the membership question never asks how many times you wrote something. So here is what the whole topic comes to. A description is a route to a set. The set is where you land. Two routes that share no words can land in the same place, and two routes that look almost identical can land somewhere different by a minus sign.

12:58Which is why equality is never settled by comparing descriptions. It is settled by comparing members, in both directions, with nothing skipped. And when they are not equal, one member says so. You do not owe anybody the full list. Descriptions are disposable. Membership is not. Here is one to try. Find two descriptions that share no words and name the same set, then prove it - both directions, out loud.

13:31The proof is the part that counts. Tell me what you came up with.